INEQUALITY'S BLOG

November 20, 2010

Problem 59: van khea

Filed under: Uncategorized — KKKVVV @ 9:12 am

Let a, b, c>0 ; ab+bc+ca=1. Prove that:
\displaystyle \frac{a}{\sqrt[3]{1+a^2}}+\frac{b}{\sqrt[3]{1+b^2}}+\frac{c}{\sqrt[3]{1+c^2}}\leq \sqrt[3]{\frac{9}{4}(a+b+c)}

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